f2+20f=44

Solution for f2+20f=44 equation:

f2+20f=44

We move all terms to the left:

f2+20f-(44)=0

We add all the numbers together, and all the variables

f^2+20f-44=0

a = 1; b = 20; c = -44;
Δ = b2-4ac
Δ = 202-4·1·(-44)
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-24}{2*1}=\frac{-44}{2}
=-22
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+24}{2*1}=\frac{4}{2}
=2
$